Question: The velocity of a particle moving along the $x$ -axis is $v(t)=\dfrac{1}{\sqrt{t}}$. At $t=4$, its position is $2$. What is the position of the particle, $s(t)$, at any time $t$ ? Choose 1 answer: Choose 1 answer: (Choice A) A s ( t ) = 2 t 1 2 s(t)=2t\^{\frac{1}{2}} (Choice B) B s ( t ) = 2 t 1 2 − 2 s(t)=2t\^{\frac{1}{2}}-2 (Choice C) C s ( t ) = 2 3 t 3 2 s(t)=\dfrac{2}{3}t\^{\frac{3}{2}} (Choice D) D s ( t ) = 2 3 t 3 2 − 10 3 s(t)=\dfrac{2}{3}t\^{\frac{3}{2}} -\dfrac{10}{3}
Solution: We know that $s(t)= \int v(t) \,dt$. In this case, $s(t)= \int \dfrac{1}{\sqrt{t}} \,dt$ Let's find the indefinite integral: ∫ 1 t √ d t = 2 t 1 2 + C \begin{aligned} \int \dfrac{1}{\sqrt{t}} \,dt&=2t\^{\frac{1}{2}}+C\\ \end{aligned} We know that $s(4)=2$. Let's use this information to solve for $C$. s ( t ) s ( 4 ) 2 − 2 = 2 t 1 2 + C = 2 ( 4 ) 1 2 + C = 4 + C = C \begin{aligned}s(t)&=2t\^{\frac{1}{2}}+C\\ \\ s(4)&=2(4)\^{\frac{1}{2}}+C\\ \\\\ 2&=4+C\\ \\ -2&=C \end{aligned} The position of the particle $s(t)$ at any time $t$ is s ( t ) = 2 t 1 2 − 2 s(t)=2t\^{\frac{1}{2}}-2.